Question

A point about a scalar multiple and a square-matrix determinant.

Answer 1

Let \(\color{#000000}{ \displaystyle {\rm A} _1 }\) by (a non-singular) \(\color{#000000}{ \displaystyle {\rm n}\times {\rm n} }\) size matrix, and let \(\color{#000000}{ \displaystyle {\rm A}_2={\rm c}\cdot {\rm A}_1 }\), then \(\color{#000000}{ \displaystyle \left|{\rm A}_2\right|={\bf c}^{\rm n}\left|{\rm A}_1\right| }\).

Answer 2

Could be easily sensed, using the Co-factor Expansion, because you will multiply times \(\color{#000000}{ \displaystyle {\bf c} }\), \(\color{#000000}{ \displaystyle {\rm n} }\) number of times.

Answer 3

Just Sharing:)

Answer 4

yes!
you can factor out c from each of the n columns or rows, then take a determinant :)

Answer 5

I was playing with Mathematica, and accidentally came across this. First I tried c=10, and particular cases, then recalled Co-factor Expansion tech. Yes, it might be easier to take \(c^n\) out, but not in the problems that instructors assign.